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I read from wikipedia that the geometric standard deviation is

$$\sigma_g = \text{exp}\left( \sqrt{\frac{\sum_{i=1}^n\left(\frac{A_i}{\mu_g}\right)^2}{n}}\right)\,, $$

where $\mu_g$ is the geometric mean, $A_i$ is the observation $i$, and $n$ is the sample size.

How can I calculate a confidence interval for a geometric mean?

Is the standard error for the geometric mean $se_g = \frac{\sigma_g}{\sqrt{n}}$? Assuming the distribution of $A$ is normal, is my 95% confidence interval for the geometric mean $\mu_g - 1.96 \cdot se_g < \mu_g < \mu_g + 1.96 \cdot se_g$

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  • $\begingroup$ Remi.b -- on reading the link, you have not correctly described the formula. At no place there is it called "the standard deviation for the geometric mean", which is highly misleading (suggesting it's the standard deviation of the geometric mean). I've corrected your question and I'll delete my answer, since (after that correction) it's now a duplicate as you indicate. $\endgroup$ – Glen_b -Reinstate Monica Jun 7 '15 at 0:06
  • $\begingroup$ I still have an issue but I'll address it in a new post. This post is a duplicate and I vote to close. Thank you for your help @Glen_b $\endgroup$ – Remi.b Jun 7 '15 at 0:09